Fermat's theorem

Note that such a number cannot be divisible by the square of a prime $p$ , because it wouldn't divide $p^{25}-p$ . So we're only looking at primes and products of distinct primes.

Which primes divide $a^{25} - a$ for all $a$ ? If $g$ is a primitive root mod $p$ , then $g^{24} \equiv 1$ mod $p$ implies that $(p-1) | 24$ . This happens for $p =2,3,5,7,13$ . And Fermat's little theorem implies that all of those primes do divide $a^{25}-a$ for all $a$ .

Any product of these primes also has that property. There are $32$ positive numbers that are products of those five distinct primes, and $\fbox{31}$ of them are $> 1$ .